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%I #14 Dec 12 2022 10:19:36
%S 0,1,1,1,0,0,2,2,2,2,2,3,-11,0,0,2,2,2,3,3,3,16,0,0,2,3,3,3,3,3,
%T 2220422932,0,0
%N Sum of three cubes problem: a(n) = integer x with the least possible absolute value such that n = x^3 + y^3 + z^3 with |x| >= |y| >= |z|, or 0 if no such x exists.
%C It is known that there is no solution for n congruent to +-4 (mod 9), but it is now conjectured that there is a solution (and probably infinitely many such) for all other numbers. The numbers n = 0, 1 and 2 are the only cases for which infinite families of parametric solutions are known, for other n the solutions seem to be sporadic.
%C Search on this problem was motivated by a statement in Mordell's paper from 1953. Beck et al. found a solution for n = 30 in 1999, and for 52 in 2000. Huisman found a solution for n = 74 in 2016. A solution for 33 was found by Booker in 2019. The number 42 was the last one below 100 for which a solution was found, in late 2019, using a collaborative effort with supercomputers and home computers from volunteers.
%C For n < 30, we have a(n) = A246869(n+1) for the nonzero values, while A246869(n+1) = 2 for n == 4 or 5 (mod 9) up to there.
%H Michael Beck, Eric Pine, Wayne Tarrant and Kim Yarbrough Jensen, <a href="https://doi.org/10.1090/S0025-5718-07-01947-3">New integer representations as the sum of three cubes</a>, Math. Comp. 76 (2007) pp. 1683-1690, DOI:10.1090/S0025-5718-07-01947-3.
%H A. R. Booker, <a href="https://doi.org/10.1007/s40993-019-0162-1">Cracking the problem with 33</a>, Res. Number Theory 5 no. 26 (2019), DOI:10.1007/s40993-019-0162-1.
%H V. L. Gardiner, R. B. Lazarus and P. R. Stein: <a href="http://DOI.org/10.2307/2003763">Solutions of the Diophantine Equation x^3 + y^3 = z^3 - d</a>, Math.Comp. 18, No. 87 (1964), pp. 408-413. DOI: 10.2307/2003763.
%H Jon Grantham and P.G. Walsh, <a href="https://arxiv.org/abs/2211.12149">Representing integers as a sum of three cubes</a>, arXiv preprint (2022). arXiv:2211.12149 [math.NT]
%H B. Haran, <a href="https://www.youtube.com/playlist?list=PLt5AfwLFPxWJcqG5YM89Qes5gZdAFM4Q1">Sum of Three Cubes - Numberphile</a>, YouTube playlist (featuring videos from Nov 06 2015, May 31 2016 (about 74), Mar 12 2019 (about 33), Sep 2019 (about 42).
%H Sander G. Huisman, <a href="http://arXiv.org/abs/1604.07746">Newer sums of three cubes</a>, arXiv:1604.07746 [math.NT] (2016).
%H J. C. P. Miller, M. F. C. Woollett, <a href="https://doi.org/10.1112/jlms/s1-30.1.101">Solutions of the Diophantine Equation x^3+y^3+z^3=k, Journal of the London Mathematical Society, s1-30 (1955) pp. 101-110. DOI: 10.1112/jlms/s1-30.1.101.
%H L. J. Mordell, <a href="https://doi.org/10.1112/jlms/s1-28.4.500">Integer Solutions of the Equation x^2+y^2+z^2+2xyz = n, Journal London Math. Soc. s1-28 no. 4 (1953) pp. 500‐510.
%H Bjorn Poonen, <a href="https://www.ams.org/journals/notices/200803/tx080300344p.pdf">Undecidability in number theory</a>, Notices Amer. Math. Soc. 55 (2008), no. 3, pp. 344-350.
%F a(n) = 0 for n == 4 or n == 5 (mod 9).
%F a(n) <= k if |n - k^3| < 3 or |n - 2*k^3| < 2 or n = 3*k^3 for some k.
%F a(n) = A246869(n+1) for all n < 30 with a(n) > 0.
%e 0 = 0^3 + 0^3 + 0^3, 1 = 1^3 + 0^3 + 0^3,
%e 2 = 1^3 + 1^3 + 0^3, 3 = 1^3 + 1^3 + 1^3,
%e 6 = 2^3 - 1^3 - 1^3, 7 = 2^3 - 1^3 + 0^3,
%e 8 = 2^3 + 0^3 + 0^3, 9 = 2^3 + 1^3 + 0^3,
%e 10 = 2^3 + 1^3 + 1^3, 11 = 3^3 - 2^3 - 2^3,
%e 12 = -11^3 + 10^3 + 7^3, 15 = 2^3 + 2^3 - 1^3,
%e 16 = 2^3 + 2^3 + 0^3, 17 = 2^3 + 2^3 + 1^3,
%e 18 = 3^3 - 2^3 - 1^3, 19 = 3^3 - 2^3 + 0^3,
%e 20 = 3^3 - 2^3 + 1^3, 21 = 16^3 - 14^3 - 11^3,
%e 24 = 2^3 + 2^3 + 2^3, 25 = 3^3 - 1^3 - 1^3,
%e 26 = 3^3 - 1^3 + 0^3, 27 = 3^3 + 0^3 + 0^3,
%e 28 = 3^3 + 1^3 + 0^3, 29 = 3^3 + 1^3 + 1^3,
%e 30 = 2220422932^3 - 2218888517^3 - 283059965^3 was discovered by Beck, Pine, Yarbrough and Tarrant in 1999 following an approach suggested by N. Elkies.
%e 33 = 8866128975287528^3 - 8778405442862239^3 - 2736111468807040^3 was found by A. Booker in 2019. It is uncertain whether these are the smallest solutions.
%o (PARI) apply( A332201(n,L=oo)={!bittest(48,n%9)&& for(c=0,L, my(t1=c^3-n, t2=c^3+n, a); for(b=0,c,((ispower(t1-b^3,3,&a)&&abs(a)<=c)||(ispower(t1+b^3,3,&a)&&abs(a)<=c))&&return(c); ispower(t2-b^3,3,&a) && abs(a)<=c && return(-c)))}, [0..29])
%Y Cf. A060464, A060465, A060466, A060467, A246869.
%K sign,more,hard
%O 0,7
%A _M. F. Hasler_, Feb 08 2020
%E a(31) = a(32) = 0 added by _Jinyuan Wang_, Feb 15 2020
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